Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 17600 N for 55 ms.

0.8 m/s

Explanation:

F(avg) * Δt = I, where

F(avg) = Average force

Δt = change in time

I = impulse

From the question, we know that the

average force is given as -17600 N

time change is given as 55 milliseconds

speed of the player, v = 8 m/s

Mass of the player is given as 110 kg

Impulse, I = F(avg) * Δt

I = -17600 * 0.055

I = -968

Now, using the Impulse-Momentum theory, we have that

Δp = I and thus,

Δp = m [v(f) - v(i)], where

Δp = change in the momentum

v(f) = final speed of the player

v(i) = initial speed of the player

Substituting the values, we have

I = m [v(f) - v(i)]

-968 = m.v(f) - m.v(i)

m.v(f) = m.v(i) - 968

110v(f) = 110 * 8 - 968

110v(f) = 880 - 968

110v(f) = -88

v(f) = -88 / 110

v(f) = -0.8 m/s

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explanation:

d.7.5cm is your answer